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PS: Catch up Speed [#permalink]
 22 Apr 2009, 17:49
Question Stats:
54% (02:09) correct
45% (00:06) wrong based on 3 sessions
Andrew and Stephen drive on the highway in the same direction at respective rates of 72 and 80 kmh. If Stephen is 4 km behind Andrew, by how much does he have to increase his speed to catch up with Andrew in 20 minutes?
1 kmh
2 kmh
3 kmh
4 kmh
5 kmh
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Re: PS: Catch up Speed [#permalink]
22 Apr 2009, 18:44
A -- 72 Kmh
B -- 80 Kmh
Relative speed = 8 kmh
distance b/w them = 4 Km
4 /( 80 +x - 72) = 20/60
solve for x = 4 .....
Ans . D
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Re: PS: Catch up Speed [#permalink]
02 May 2009, 07:36
If x is the distance travelled by Andrew in 20 minutes = 72*20/60 = 24km Stephen's speed would to travel (24+4)km = 28*60/20 = 84 km/h So, answer is D - 4km/h
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Re: PS: Catch up Speed [#permalink]
02 May 2009, 15:25
relative speed=80-72=8
need to catch up 4km
currently Stephen is catching up at a speed of 4km/30mins, or 2km/15mins
he needs to increase the speed, so that he can make 4km in 20 mins instead, or 1km/5mins
x - the relative speed increase we seek
2/15+x=1/5
x=1/15
So, he needs to increase relative speed by 1km per 15mins, or 4km/h
Therefore, the answer is D.
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Re: PS: Catch up Speed [#permalink]
25 Jul 2009, 13:42
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I think its much easier to first assume that they are going the same speed. So he has to catch up 4 meters in 1/3 hour. r= d/t = 4/(1/3) = 12 km/hr faster he has to go. He is already going 8km/h faster, so 12 - 8 = 4 km/hr
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Re: PS: Catch up Speed [#permalink]
25 Jul 2009, 13:53
Hey guys, look the solution of benpack03. IT IS VERY FAST.
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Re: PS: Catch up Speed [#permalink]
25 Jul 2009, 14:40
benpack03 wrote: I think its much easier to first assume that they are going the same speed. So he has to catch up 4 meters in 1/3 hour. r= d/t = 4/(1/3) = 12 km/hr faster he has to go. He is already going 8km/h faster, so 12 - 8 = 4 km/hr Yes! That's what I also came up with.
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Re: PS: Catch up Speed
[#permalink]
25 Jul 2009, 14:40
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